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ANALYTICAL RESEARCHES CONNECTED WITH 
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we have ¿> 3 = /3 2 + 7 2 — 4s (7 + </>/3 ) /3 
S' 2 = /3' 2 + 7 ' 2 -4s( 7 7 + <£/3 7 )/3 7 
S" 2 = /3" 2 + 7 //2 - 4s (7" + </>/T) /3", 
and a, a 7 , a 77 will be supposed henceforth to satisfy these equations. 
We have next 
A' 2 + B' 2 -G' 2 =4( 7 7 + </>/3')/3 7 (s + cj> + Z + sZ 2 ) 
xl" 2 + 5-2 _ c»2 = 4 ( 7 " + <£/3") £"(s + + Y + sY 2 ), 
which may be simplified by writing 
fj,-<f> 1 +p<f> 
l+ja 2 ’ V /x-^’ 
where /x, v are to be considered as given functions of s and (f>. These values give 
A' 2 + B' 2 -C’ 2 = 4 ( 7 7 +(f>/3')/3's (Z + /jl)(Z + v), 
A"* + B 772 - G" 2 = 4 (7" + <¿>/3") /3 77 s(Y +/j)(Y+ v). 
Hence, putting for simplicity 
¿ 2 = 4 ( 7 7 + <f>/3') (7" + <£/3 77 ) /3 7 /3 7 ', 
we have 
4 (Z + ¿x)(Z + v)(Y+ fi) (Y+ v) = U 2 + k [(a + /3 (Y+ Z) + yYZ) 2 -8 2 (1 + F 2 ) (1 + Z 2 )\; 
and the two sides have next to be expressed in terms of Y + Z and YZ. 
If for symmetry we write 
f = l, v = Y+Z, £=YZ, 
4 (/x 2 f + [Mi + £) (y 2 £ + vr/ 4- £) + &S 2 [(| — £) 2 + t? 2 ] = U 2 +Jc (tx% + /3?7 + 7^) 2 ; 
and £7 is now to be considered a linear function of £, 7;, £ 
The condition that the first side of the equation may divide into factors, gives an 
equation for determining k; since the condition is satisfied for k = 0 and k — oc, the 
equation will be linear, and it is easily seen that the value is k = (/x — v) 2 . In fact 
4 (/¿ 2 £ + m + 0 0 2 £ + vy + 0 2 + O - v f [(£ - £) 2 + v 2 ] 
= (2/xv£ + (n+v)r} + 2£) 2 +(j*- v) 2 (f + £) 2 ; 
{2/a.f + ( tl + v )r,+ 2?)* - № = {(«f + /3, + 7 ?) ! - S’ (f + O’),